All categories

2 years ago

How many atoms of Na are in 1.89 mol of Na?

Chemistry

1 answer:

atroni [7]2 years ago

5 0

Answer:

1.138158E24 atoms or 1.14 x 10^24 atoms

Explanation:

To find atoms/particles from moles you just want to convert using avogadro's number which is 6.022 x 10^23

1.89 mol x 6.022 • 10^23

———— = 1.138158E24 atoms

1 mol

so 1.138158E24 atoms or 1.14 x 10^24 for scientific notation

hope this helps :)

You might be interested in

Solid barium sulfate dissolves into its respective ions at 25°C. Suppose that in a particular solution, [Ba2+ ] = 1.76 x 10-3 M.

ale4655 [162]

BaSO₄(s) ⇄ Ba²⁺(aq) + SO₄²⁻(aq)

Ksp=[Ba²⁺][SO₄²⁻]

[Ba²⁺]=[SO₄²⁻]

Ksp=[Ba²⁺]²

Ksp=(1.76*10⁻³)² =3.0976×10⁻⁶ ≈3.1×10⁻⁶

5 0

3 years ago

Read 2 more answers

28 ml of 0.10 m hcl is added to 60 ml of 0.10 m sr(oh)2. determine the concentration of oh− in the resulting solution.

Blababa [14]

Here is an acid-base reaction. Hydrochloric acid (HCl) reacts with strontium hydroxide [ Sr(OH)2 ]

Ions H+ and OH- neutralize each other. If the amounts are not equal, one of them will be in excess.
Follow the steps as

1. Find moles of ions: mole= Molarity * Volume (in liter) ; n= M * V OR millimole = Molarity * Volume (in ml) ;
2. Write the equation
3. Find out excess ion
4. Use final volume (V acid + V base ) to calculate concentration of excess ion.

n HCI = 28 ml * 0.10 M = 0.28 mmol, releases 0.28 mmol H+ ions
n Sr(OH)2= 60 ml * 0.10 M= 0.60 mmol, releases 2* 0.60=1.20 mmol OH- ions
since Sr(OH)2⇒ Sr2+ + 2OH-

Neutralization reaction is OH- + H+ ---> H2O. The ratio is 1:1. That means 1 mmol hydroxide ions will neutralize 1 mmol hydrogen ions. Since OH- ions are greater in amount, they will be in excess

n(OH-) - n(H+)= 1.20 - 0.28 = 0.92 mmol OH- ions UNREACTED.

Total volume= V acid + V base= 28 ml + 60 ml = 98 ml

Molarity of OH- ions= mole / Vtotal = 0.92/98= 0.009 M

The answer is 0.009 M.

5 0

3 years ago

Read 2 more answers

What is the balanced chemical equation for the combustion of sulphur

nikitadnepr [17]

Answer:

S+O2⇒SO2

Explanation:

3 0

2 years ago

Read 2 more answers

A buffer is prepared by adding 300. 0 ml of 2. 0 mnaoh to 500. 0 ml of 2. 0 mch3cooh. what is the ph of this buffer? ka= 1. 8 10

Anton [14]

The Henderson-Hasselbalch equation can be used to determine the pH of the buffer from the pKa value. The pH of the buffer will be 4.75.

<h3>What is the Henderson-Hasselbalch equation?</h3>

Henderson-Hasselbalch equation is used to determine the value of pH of the buffer with the help of the acid disassociation constant.

Given,

Acid disassociation constant (ka) = 1. 8 10⁻⁵

Concentration of NaOH = 2.0 M

Concentration of CH₃COOH = 2.0 M

pKa value is calculated as,

pKa = -log Ka

pKa = - log (1. 8 x 10⁻⁵)

Substituting the value of pKa in the Henderson-Hasselbalch equation as

pH = - log (1. 8 x 10⁻⁵) + log [2.0] ÷ [2.0]

pH = - log (1. 8 x 10⁻⁵) + log [1]

= 4.745 + 0

= 4.75

Therefore, 4.75 is the pH of the buffer.

Learn more about the Henderson-Hasselbalch equation here:

brainly.com/question/27751586

#SPJ4

6 0

1 year ago

How to change 5 % W/V of NaCl to ppm , M ? molar mass = 58.5<br>please clear explain

11Alexandr11 [23.1K]

Answer:

50000ppm and 0.855M.

Explanation:

ppm is an unit of chemistry defined as the ratio between mg of solute (NaCl) and Liters of solution. Molarity, M, is the ratio between moles of NaCl and liters

A 5% (w/v) NaCl contains 5g of NaCl in 100mL of solution.

To solve the ppm of this solution we need to find the mg of NaCl and the L of solution:

5g * (1000mg / 1g) = 5000mg

<em>L Solution:</em>

100mL * (1L / 1000mL) = 0.100L

ppm:

5000mg / 0.100L = 50000ppm

To find molarity we need to obtain the moles of NaCl in 5g using its molar mass:

5g * (1mol / 58.5g) = 0.0855moles NaCl

Molarity:

0.0855mol NaCl / 0.100L = 0.855M

7 0

3 years ago